Is this the correct reason that the sequence cannot converge to zero?

Question

Let the sequence $\left\{a_n\right\}$ is defined as: $$a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}$$

Basically the terms of the sequence are $\frac{1}{2},\frac{7}{12},\frac{37}{60},...$

Now we need to find $\lim_{n \to \infty}a_n$. It is pretty obvious that its a Riemann summation and we get the answer as $\ln 2$.

But if a novice solves this, he/she will take individual limits as follows: $$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{1}{n+1}+\lim_{n \to \infty}\frac{1}{n+2}+....+\lim_{n \to \infty}\frac{1}{2n}=0+0+0...+0=0$$

But if we observe that: $$a_{n+1}-a_{n}=\frac{1}{n+2}+\cdots+\frac{1}{2n+2}-\frac{1}{n+1}-\cdots-\frac{1}{2n}=\frac{1}{2n+1}-\frac{1}{2n+2}>0$$ So the sequence $\left\{a_n\right\}$ is increasing and bounded below by $\frac{1}{2}$.So incase it converges it converges to a value more than $\frac{1}{2}$.Hence the limit $0$ is the wrong answer.Is this reasoning valid or can we reason in a better for a novice calculus student?

Answer 1

For me, the simplest approach consists in observing that each term in the sum is $\ge\frac 1{2n}$, and that there are $n$ terms. Therefore $$a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{n+n}\ge n\cdot\frac1{2n}=\frac12.$$

Answer 2

Well, the thing to note is that in the limit, "you are summing up infinitely many $0$s".

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To convince a novice that the argument is not correct, you can give an even simpler example. Consider $$a_n = \underbrace{\frac{1}{n} + \cdots + \frac{1}{n}}_{n \text{ times}}.$$ By the same argument, you would get $a_n \to 0$ but in fact, we have $a_n = 1$ for all $n \ge 1$.

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To convince them that the limit in your particular case is not $0$, I think what you have written is a simple enough reason.

Edit: See Bernard's answer.